# Compact Set in Metric Space

by wayneckm
Tags: compact, metric, space
 P: 68 Hello all, Here is my question while reading a proof. For a compact set $$K$$ in a separable metrizable spce $$(E,\rho)$$ and a continuous function $$t \mapsto f(t)$$, if we define $$D_{K} = \inf \{ t \geq 0 \; : \; f(t) \in K \}$$ then, $$D_{K} \leq t$$ if and only if $$\inf\{ \rho(f(q),K) : q \in \mathbb{Q} \cap [0,t] \}$$ = 0 May someone shed some light on this? I do not understand it. Thanks very much. Wayne
 P: 68 Domain of $$f$$ is $$\mathbb{R}^{+}$$ Codomain of $$f$$ is $$\mathbb{R}$$
 Sci Advisor P: 905 Compact Set in Metric Space Huh? Then I don't understand $f(t)\in K$...
 P: 68 Oops...sorry, i misunderstood the term codomain. So codomain here should be $$E$$ as stated.
 Emeritus Sci Advisor PF Gold P: 4,500 If $$D_k \leq t$$ then $$f(D_k)\in K$$. We can approximate $$D_k$$ with rational numbers, and because $$D_k \in [0,t]$$ we can approximate $$D_k$$ with rational numbers in $$\mathbb{Q}\cap [0,t]$$ If $$q_r$$ is such a sequence converging to $$D_k$$, the distance between $$f(q_r)$$ and $$f(D_k)$$ goes to zero, which means the distance between $$f(q_r)$$ and $$K$$ must go to zero. So the infimum of the distance between $$f(q)$$ and $$K$$ for $$q\in \mathbb{Q}\cap [0,t]$$ must be zero because we just found a sequence for which the distance is arbitrarily small. This is basically the direction $$D_k\leq t$$ implies the infimum is zero.